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Finding Fourier Series Coefficients

Module by: Don Johnson

Summary: This module outlines the procedure for determining Fourier series coefficients.

Assume for the moment that the Fourier series works. To find the Fourier coefficients, we note the following orthogonality properties of sinusoids.

Orthogonality

∀k,l:∫0Tsin2πktTcos2πltTdt=0 k l t 0 T 2 k t T 2 l t T 0 (1)
∫0Tsin2πktTsin2πltTdt=T2ifk=l∧k≠0∧l≠00ifk≠l∨k=0=l t 0 T 2 k t T 2 l t T T 2 k l k 0 l 0 0 k l k 0 l (2)
∫0Tcos2πktTcos2πltTdt=T2ifk=1∧k≠0∧l≠0Tifk=0=l0ifk≠l t 0 T 2 k t T 2 l t T T 2 k 1 k 0 l 0 T k 0 l 0 k l (3)
To use these, let's multiply the Fourier series for a square wave ( st= a 0 +∑k=1∞ a k cos2πktT+∑k=1∞ b k sin2πktT s t a 0 k 1 a k 2 k t T k 1 b k 2 k t T ) by cos2πltT 2 l t T and integrate. The idea is that, because integration is linear, the integration will sift out all but the term involving a l a l .
∫0Tstcos2πltTdt=∫0T a 0 cos2πltTdt+∑k=1∞ a k ∫0Tcos2πktTcos2πltTdt+∑k=1∞ b k ∫0Tsin2πktTcos2πltTdt t 0 T s t 2 l t T t 0 T a 0 2 l t T k 1 a k t 0 T 2 k t T 2 l t T k 1 b k t 0 T 2 k t T 2 l t T (4)
The first and third terms are zero; in the second, the only non-zero term in the sum results when the indices kk and ll are equal (but not zero), in which case we obtain a l T2 a l T 2 . If k=0=l k 0 l , we obtain a 0 T a 0 T . Consequently, ∀l,l≠0: a l =2T∫0Tstcos2πltTdt l l 0 a l 2 T t 0 T s t 2 l t T All of the Fourier coefficients can be found similarly.
a 0 =1T∫0Tstdt a 0 1 T t 0 T s t (5)
∀k,k≠0: a k =2T∫0Tstcos2πktTdt k k 0 a k 2 T t 0 T s t 2 k t T (6)
b k =2T∫0Tstsin2πktTdt b k 2 T t 0 T s t 2 k t T (7)

Exercise 1

The expression for a 0 a 0 is referred to as the average value of st s t . Why?

Solution 1

The average of a set of numbers is the sum divided by the number of terms. Viewing signal integration as the limit of a Riemann sum, the integral corresponds to the average.

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